$$ y = \operatorname*{argmin}_\hat{x} \operatorname*{E}_x L(|\hat{x} - x|) $$

where $L$ is a loss function. As noted here, if $f(s) = s^p$ then

  • $p \rightarrow 0$ implies $y \rightarrow$ the mode
  • $p = 1$ implies $y =$ the median
  • $p = 2$ implies $y =$ the mean
  • $p \rightarrow \infty$ implies $y \rightarrow$ the midrange

Is there an $L$ bounded on $[0,\infty)$ whose minimizer $y$ is also the mean?

  • $\begingroup$ My gut: not without some constraints on the range of $x$ or its residuals. And cutting the loss function off at the biggest possible $x-\hat{x}$ seems like a cheater's answer. $\endgroup$ – AdamO Oct 23 '19 at 16:14

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